Prerequisites. We recall: Theorem 2 A subset of a countably innite set is countable.

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1 Prerequisites 1 Set Theory We recall the basic facts about countable and uncountable sets, union and intersection of sets and iages and preiages of functions. 1.1 Countable and uncountable sets We can copare innite sets via bijections or one-to-one correspondences. Denition 1 Let I be an arbitrary set. a) The set I is nite, if there is a bijective ap f : I {1, 2, 3,..., n} for soe positive integer n. b) The set I is innite, if it is not nite. c) The set I is countably innite, if there is a bijective ap f : I N. d) The set I is countable, if it is either nite or countably innite. e) The set I is uncountable, if it is not countable. We recall: Theore 2 A subset of a countably innite set is countable. We have furtherore the iportant theore: Theore 3 A countable union of countable sets is countable. proof: It is sucient to prove the stateent for a disjoint union A = i=1 A i of countably innte sets A i. This is true as 1.) Each union of sets i=1 B i can be decoposed into a disjoint union reoving ultiple occurences. i=1 B i of sets by 2.) Each nite set B i can be extended to an innite set B i, such that B i B k = for all k i. 3.) If i=1 B i is countably innite, then the subset i=1 B i is countable by Theore 2.

2 So suppose we have a disjoint union i=1 A i of countably innte sets A i. We list all eleents of A = i=1 A i: A 1 = {x 11, x 12, x 13,..., x 1n,...} A 2 = {x 21, x 22, x 23,..., x 2n,...}. A = {x 1, x 2, x 3,..., x n,...}. As the factorization into pries is unique, we know that the set of positive integers S = {2 k 3 n, n, k N} satises: 2 k1 3 n 1 = 2 k2 3 n 2 k 1 = k 2 and n 1 = n 2. (*) Hence the assignent f : S i=1 A i, dened by f(2 k 3 n ) = x kn is a well-dened ap which is bijective. Hence i=1 A i is in one-to-one correspondence with a subset of N, which by Theore 2 is countable. Hence A = i=1 A i is also countable. Exaples 4 N, Z and Q are countable, hence by the previous theore we know that Z 2 = Z Z = (i, Z) and Q 2 = Q Q = (q, Q) i Z q Q are countable. Using this arguent iteratively we have that for xed n, Z n and Q n are countable. 1.2 Sets and functions Theore 1 (De Morgan's Law) Let (A i ) X be a collection of sets in X. If A c = X\A for all A X, then a) ( A ) c i = Ac i b) ( A ) c i = Ac i proof a) We have ( ) c A i = X\{x X i I, such that x A i } = {x X ( i I, such that x A i )} = {x X i I, x A i } = {x X i I, x A c i} = A c i.

3 b) Siilarly ( ) c A i = X\{x X i I, x A i } = {x X ( i I, we have that x A i )} = {x X i I, such that x A i } = {x X i I, x A c i} = A c i. Lea 2 (Functions and Sets) Let f : X Y be a function. Let (A j ) j J X be a collection of sets in X. Let furtherore (B i ) Y be a collection of sets in Y and B Y. Then ( ) a) f j J A j j J f (A j). b) ( ) j J f (A j) = f j J A j. c) f 1 (B) c = f 1 (B c ). d) f 1 (B i ) = f 1 ( B ) i and f 1 (B i ) = f 1 ( B i). proof a) We have that f A j = {y Y y = f(x) and f(x) f( A j )} = j J j J {f(x) Y j J, we have x A j }. Furtherore f (A j ) = {y Y j J, we have y f(a j )} = j J {f(x) Y j J, we have f(x) f(a j )}. Now if x A j then f(x) f(a j ) and the rst set is contained in the second. Note We see that the converse is not true by taking f : {1, 2} {1}, where f(1) = f(2) = 1. For A 1 = {1} and A 2 = {2} we get b) We know that j J f(a 1 A 2 ) =, but f(a 1 ) f(a 2 ) = {1}. f (A j ) = {y Y j J, such that y f(a j )} = {f(x) Y j J, such that f(x) f(a j )}.

4 On the other hand f A j = {y Y y = f(x) and j J, such that x A j } = j J {f(x) Y j J, such that x A j }. But if x A j then f(x) f(a j ) and the second set is contained in the rst. On the other hand if f(x) f(a j ) for soe j J then there is x, such that f(x) = f(x ) and x A j for soe j J. So the rst set is contained in the second. c) We know that f 1 (B i ) = {x X i I, such that x f 1 (B i )} = {x X i I, such that f(x) B i }. We copare this with ( ) f 1 B i = {x X f(x) B i } = {x X i I, such that f(x) B i } which shows that the sets are equal. We prove the second stateent in a siilar fashion. d) We know that f 1 (B) c = X\f 1 (B) = X\{x X f(x) B} = {x X f(x) B}. On the other hand we have that f 1 (B c ) = {x X f(x) B c } = {x X f(x) B} and the two sets are equal. 2 Topology The proofs of the following theores can be found in Munkres, Topology, 2nd edition, Chapter 2, Section 12,13 and Basics Denition 1 Let X be a set. A topology on X is a collection T P(X) of subsets of X, such that

5 a) T and X T. b) A, B T A B T (T is closed under intersection). c) (A k ) k K T k K A k T (T is closed under any union). In this case the eleents of T the open subsets of X and (X, T ) is called a topological space. Exaples T = {, X} or T = P(X). Reark 2 b) iplies that T is stable under nite intersections. Denition 3 Let (X, T ) and (X, T ) be topological spaces. A function f : X X is continuous if f 1 (A ) T for all A T. Denition 4 (Basis) Let (X, T ) be a topological space. Then β T is a basis for the topology T if for all A T we have that A = A i where (A i ) β. This eans that every eleent in T is a union of eleents of β. Theore 5 (Basis = neighbourhood basis) β is a basis for the topology T i for all A T and for all x A U(x) = U β, such that x U A. Denition 6 (second countable) A topological space (X, T ) is called second countable if there is a countable basis for its topology. Exaple A second countable basis for the usual topology of the real line R is given by the intervals with rational endpoints. Proposition 7 If (X, d) is a etric space with a countable dense subset, the topology induced by the etric is second countable. proof We know that 1.) the basis β d of the topology T d induced by the etric d is the collection of open balls in (X, d): β d = {B r (x) r R +, x X} 2.) there is a countable dense subset D = (x n ) n N X in X.

6 3.) by Theore 5, as β d is a basis, we know that for all A T d and x A there is B r (x ) β d, such that x B r (x ) A. We take β = {B 1 (x n ), n N}. Take A T and x A as in 3.). Fro this condition it follows that it is sucient to show that there is a ball B 1 (x n ) β, such that B 1 (x n ) B r (x ). Furtherore, if x x, we can nd a ball of saller radius around x that also satises 3.). Hence we can assue that x = x. To construct our ball we take N, such that r 2 > 1 r > 2. By the density of D there is x n D, such that d(x n, x) < 1. Then for every point x B 1 (x n ) we have by the triangle inequality: d( x, x) d( x, x n ) + d(x n, x) < < r Hence x B 1 (x n ) B r (x) A and therefore β is a countable basis for T. 3 Liits We recall the dention of inu and supreu and li inf and li sup. The correspondig theores and denitions can be, for exaple found in Gordon, Real Analysis - A First Course, 2nd edition. 3.1 Inu and supreu Denition 1 Let S R be a non-epty set of real nubers. Suppose S is bounded above. The nuber β is the supreu of S if β is an upper bound of S and any nuber less than β is not an upper bound of S i.e. We will write β = sup(s). for all b < β there is an x S, such that b < x. Denition 2 Let S R be a non-epty set of real nubers. Suppose S is bounded below. The nuber α is the inu of S if α is a lower bound of S and any nuber greater than α is not a lower bound of S i.e. We will write α = inf(s). for all a > α there is an x S, such that a > x.

7 3.2 The extended real nuber line see Wilkins: The extended real nuber syste. 3.3 Liit superior and liit inferior We recall the following denitions fro real analysis: Let (a n ) n N R be a sequence. For k 1 consider the new sequence b k = sup a n = sup{a k, a k+1, a k+2, a k+3,...} n k Then b k b k+1 for all k N and therefore li k b k = inf k N b k R. We dene: Denition 1 (Liit superior and inferior) We call the liit superior of a sequence (a n ) n R the nuber Def. li sup a n = li b k = inf b k. n N k k N In a siilar fashion we call the liit inferior of a sequence (a n ) n R the nuber li inf n N Exaple The sequence (a n ) n N = where c k = inf n k a n. a n Def. = li k inf n k a n. ( ) cos(n) n and the sequence (c k) k N n N Figure 1: Plot of cos(x) x (red) and the sequence (a n ) n N = given by c k = inf n k a n (blue). ( ) cos(n) n (black) and the sequence n N

8 Proposition 2 For a sequence (a n ) n N R we have that a) li inf n N a n li sup n N a n. b) li n a n exists if and only if li inf n N a n = li n a n = li sup n N a n. 4 Coplex analysis see Beck et al.: A rst course in coplex analysis.